Math Problem Statement
Let h be defined by the piecewise function h(x) = {x^2 + 1, for x < 2; kx + 1, for x ≥ 2}. For what value of k, if any, is h(x) both continuous and differentiable at x = 2?
Solution
The problem asks to find the value of such that the function , defined as:
\begin{cases} x^2 + 1 & \text{if } x < 2 \\ kx + 1 & \text{if } x \geq 2 \end{cases}$$ is both continuous and differentiable at $$x = 2$$. ### Step 1: Continuity at $$x = 2$$ For continuity at $$x = 2$$, the left-hand limit $$\lim_{x \to 2^-} h(x)$$ must equal the right-hand limit $$\lim_{x \to 2^+} h(x)$$, and both should equal $$h(2)$$. - For $$x < 2$$, $$h(x) = x^2 + 1$$. So: $$\lim_{x \to 2^-} h(x) = 2^2 + 1 = 4 + 1 = 5$$ - For $$x \geq 2$$, $$h(x) = kx + 1$$. So: $$\lim_{x \to 2^+} h(x) = 2k + 1$$ For continuity, $$\lim_{x \to 2^-} h(x) = \lim_{x \to 2^+} h(x)$$, which gives: $$5 = 2k + 1$$ Solving for $$k$$: $$2k = 4 \quad \Rightarrow \quad k = 2$$ ### Step 2: Differentiability at $$x = 2$$ For differentiability at $$x = 2$$, the derivatives from the left and right sides must be equal. - The derivative of $$h(x) = x^2 + 1$$ (for $$x < 2$$) is $$\frac{d}{dx}(x^2 + 1) = 2x$$, so: $$\lim_{x \to 2^-} h'(x) = 2(2) = 4$$ - The derivative of $$h(x) = kx + 1$$ (for $$x \geq 2$$) is $$\frac{d}{dx}(kx + 1) = k$$, so: $$\lim_{x \to 2^+} h'(x) = k$$ For differentiability, the derivatives must match: $$4 = k$$ ### Conclusion: There is **no value of $$k$$** that satisfies both the conditions of continuity and differentiability. Based on the calculations, $$k = 2$$ satisfies continuity but does not satisfy differentiability (because $$k$$ would need to be 4 for differentiability). Thus, the answer is: **(E) ThereAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Differentiability
Piecewise Functions
Limits
Formulas
Continuity condition: lim_{x→2^-} h(x) = lim_{x→2^+} h(x) = h(2)
Differentiability condition: lim_{x→2^-} h'(x) = lim_{x→2^+} h'(x)
Theorems
Continuity and Differentiability of Piecewise Functions
Suitable Grade Level
Grades 11-12
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